OCR MEI Paper 1 Specimen — Question 3 4 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
SessionSpecimen
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |linear| > constant (greater than)
DifficultyEasy -1.2 This is a straightforward application of the basic modulus inequality rule |f(x)| ≥ k splits into f(x) ≥ k or f(x) ≤ -k. Requires only direct algebraic manipulation with no problem-solving insight, making it easier than average but not trivial since students must remember to consider both cases correctly.
Spec1.02l Modulus function: notation, relations, equations and inequalities

3 Solve the inequality \(| 2 x - 1 | \geq 4\).

Question 3:
Either:
AnswerMarks Guidance
\(2x-1 \geq 4\)
\(\Rightarrow 2x - 1 \geq 4\)M1 (1.1)
or \(\quad 2x - 1 \leq -4\)M1 (1.1)
Or:
AnswerMarks Guidance
\((2x-1)^2 \geq 16\)M1 (1.1) M1 for sketch of \(y=(2x-1)^2\) and \(y=16\)
\(4x^2 - 4x - 15 \geq 0\)
AnswerMarks Guidance
\((2x-5)(2x+3) \geq 0\)M1 (1.1) M1 for \(x = 2\frac{1}{2},\ -1\frac{1}{2}\)
\(\Rightarrow x \geq 2\frac{1}{2}\)A1 (1.1)
\(\Rightarrow x \leq -1\frac{1}{2}\)
AnswerMarks Guidance
\(\{x: x \leq -1\frac{1}{2}\} \cup \{x: x \geq 2\frac{1}{2}\}\)A1 (2.5) OR \(x \geq 2\frac{1}{2}\) or \(x \leq -1\frac{1}{2}\); if final answer not in one of these forms then withhold final A1
Total: [4]
## Question 3:

**Either:**
$|2x-1| \geq 4$
$\Rightarrow 2x - 1 \geq 4$ | M1 (1.1) |
or $\quad 2x - 1 \leq -4$ | M1 (1.1) |

**Or:**
$(2x-1)^2 \geq 16$ | M1 (1.1) | M1 for sketch of $y=(2x-1)^2$ and $y=16$
$4x^2 - 4x - 15 \geq 0$
$(2x-5)(2x+3) \geq 0$ | M1 (1.1) | M1 for $x = 2\frac{1}{2},\ -1\frac{1}{2}$
$\Rightarrow x \geq 2\frac{1}{2}$ | A1 (1.1) |
$\Rightarrow x \leq -1\frac{1}{2}$
$\{x: x \leq -1\frac{1}{2}\} \cup \{x: x \geq 2\frac{1}{2}\}$ | A1 (2.5) | OR $x \geq 2\frac{1}{2}$ or $x \leq -1\frac{1}{2}$; if final answer not in one of these forms then withhold final A1
**Total: [4]**

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3 Solve the inequality $| 2 x - 1 | \geq 4$.

\hfill \mbox{\textit{OCR MEI Paper 1  Q3 [4]}}